Time and Work problems look complicated, but there’s a single trick that makes almost every problem a breeze: the LCM method. Once we learn it, we’ll wonder why anyone bothers with fractions at all.
The core idea is simple: Work = Rate × Time. If A can do a job in 10 days, then A’s rate is 1/10 of the job per day. When people work together, their rates ADD UP.
The Fraction Method (Traditional)
If A can finish work in a days, A does 1/a of the work per day. If A and B work together, combined rate = 1/a + 1/b per day. Time to finish together = 1/(1/a + 1/b) = ab/(a+b).
This works, but dealing with fractions gets messy fast. Enter the LCM method.
The LCM Method — The Game Changer
Instead of working with fractions, we assume the total work is a concrete number — specifically, the LCM of the given times. This turns everything into whole numbers. WAY faster.
Example 1: Basic LCM method
A can do a job in 12 days, B can do it in 18 days. How long to finish together?
LCM of 12 and 18 = 36 (this is our total work)
- A’s efficiency = 36/12 = 3 units/day
- B’s efficiency = 36/18 = 2 units/day
- Combined = 3 + 2 = 5 units/day
- Time = 36/5 = 7.2 days (or 7 days and 4.8 hours)
Compare with the fraction method: 1/12 + 1/18 = 3/36 + 2/36 = 5/36, time = 36/5 = 7.2 days. Same answer, but the LCM method feels more natural with whole numbers.
Example 2: Three workers
A, B, and C can do a job in 10, 15, and 20 days respectively. How long if all three work together?
LCM of 10, 15, 20 = 60 units.
- A = 60/10 = 6 units/day
- B = 60/15 = 4 units/day
- C = 60/20 = 3 units/day
- Combined = 6 + 4 + 3 = 13 units/day
- Time = 60/13 = 4 and 8/13 days (approximately 4.6 days)
Example 3: One worker leaves midway
A and B together can finish a job in 12 days. They work together for 8 days, then B leaves. A finishes the remaining work in 10 more days. Find how long A and B individually take.
Let A’s time = a days, B’s time = b days.
Together: 1/a + 1/b = 1/12 … (i)
In 8 days together, they finish 8/12 = 2/3 of the work. Remaining = 1/3.
A finishes 1/3 in 10 days, so A’s rate = (1/3)/10 = 1/30. Thus a = 30 days.
From (i): 1/30 + 1/b = 1/12 → 1/b = 1/12 - 1/30 = (5-2)/60 = 3/60 = 1/20. Thus b = 20 days.
Alternate Day Work
Sometimes A works on day 1, B works on day 2, A works on day 3, and so on.
The trick: find the combined work done in a 2-day cycle, then figure out how many complete cycles are needed.
Example 4: Alternate day work
A can do a job in 10 days, B in 15 days. They work on alternate days, starting with A. When will the work be finished?
LCM of 10 and 15 = 30 units.
- A’s efficiency = 3 units/day
- B’s efficiency = 2 units/day
In a 2-day cycle (A then B): 3 + 2 = 5 units done.
After 10 days (5 complete cycles): 5 × 5 = 25 units done. Remaining = 30 - 25 = 5 units.
Day 11 (A’s turn): A does 3 units. Remaining = 5 - 3 = 2 units.
Day 12 (B’s turn): B does 2 units. Remaining = 0. Done!
Work finishes on day 12.
Pipes and Cisterns
Think of it like this: inlet pipes do “positive work” (filling the tank), and outlet pipes/leaks do “negative work” (emptying the tank). Everything else is the same as time and work.
Example 5: Pipes problem
Pipe A fills a tank in 12 hours, pipe B fills it in 15 hours, and pipe C empties it in 20 hours. If all three are opened, how long to fill the tank?
LCM of 12, 15, 20 = 60 units.
- A fills: 60/12 = +5 units/hour
- B fills: 60/15 = +4 units/hour
- C empties: 60/20 = -3 units/hour
Net rate = 5 + 4 - 3 = 6 units/hour
Time = 60/6 = 10 hours
Example 6: Leak problem
A pipe fills a tank in 6 hours. Due to a leak, it takes 8 hours. How long would the leak alone take to empty the full tank?
LCM of 6 and 8 = 24 units.
- Pipe rate = 24/6 = 4 units/hour
- Net rate (pipe + leak) = 24/8 = 3 units/hour
- Leak rate = 4 - 3 = 1 unit/hour (draining)
Time for leak to empty = 24/1 = 24 hours
Efficiency-Based Problems
“A is twice as efficient as B” means A’s rate is 2 times B’s rate. If B takes 20 days, A takes 10 days.
“A is 50% more efficient than B” means A’s rate is 1.5 times B’s rate. If B takes 30 days, A takes 30/1.5 = 20 days.
Example 7: Efficiency problem
A is 60% more efficient than B. If B can do a job in 40 days, how long do they take together?
A’s time = 40/1.6 = 25 days.
LCM of 25 and 40 = 200 units.
- A = 200/25 = 8 units/day
- B = 200/40 = 5 units/day
- Combined = 13 units/day
- Time = 200/13 = 15 and 5/13 days (approximately 15.38 days)
Common Exam Variations
- A and B work together, one leaves partway — find individual times.
- Alternate day work — who finishes and on which day?
- Pipes and cisterns with multiple inlets and outlets.
- “A does half the work, then B finishes the rest.”
- Efficiency ratios: “A is x times as efficient as B.”
- “If 5 men can do it in 10 days, how many men needed for 5 days?” (Use: Men × Days = constant)
- MDH formula: M₁ × D₁ × H₁ = M₂ × D₂ × H₂ (Men × Days × Hours = constant work)
The MDH Formula
When the work stays the same:
M₁ × D₁ × H₁ = M₂ × D₂ × H₂
This means: if 5 men working 8 hours/day finish in 10 days, how many men are needed working 10 hours/day to finish in 4 days?
5 × 10 × 8 = M₂ × 4 × 10 → M₂ = 400/40 = 10 men
Practice Problems
Problem 1: A can do a job in 20 days, B in 30 days. They start together, but A leaves after 5 days. How many more days does B take to finish?
Problem 2: Two pipes can fill a tank in 10 and 12 hours. A drain pipe can empty it in 20 hours. If all three are opened, how long to fill the tank?
Problem 3: 12 men can finish a job in 16 days working 8 hours/day. How many men are needed to finish the same job in 12 days working 10 hours/day?
Answers
Problem 1: LCM of 20 and 30 = 60 units. A = 3/day, B = 2/day. In 5 days together: 5 × (3+2) = 25 units done. Remaining = 60 - 25 = 35 units. B alone: 35/2 = 17.5 more days.
Problem 2: LCM of 10, 12, 20 = 60 units. Pipe 1 = +6, Pipe 2 = +5, Drain = -3. Net = 8 units/hour. Time = 60/8 = 7.5 hours.
Problem 3: M₁D₁H₁ = M₂D₂H₂ → 12 × 16 × 8 = M₂ × 12 × 10 → M₂ = 1536/120 = 12.8, so 13 men (round up since we can’t have partial men).